How to show $\mathbb{E}[AB\mid B]=B\cdot\mathbb{E}[A\mid B]$? How to show $\mathbb{E}[AB\mid B]=B\cdot\mathbb{E}[A\mid B]$?
Intuitively, since we are conditioning on $B$, $B$ is already known so we can simply take $B$ out of the expectation operator. But the tricky part is $\mathbb{E}[AB\mid B]$ is a random variable.
 A: I will use the following definition as my starting point:

Definition. If $X \in L^1(\Omega, \mathcal{F}, \Bbb{P})$ and $\mathcal{G} \subseteq \mathcal{F}$ is a $\sigma$-algebra, then $\Bbb{E}[X \mid \mathcal{G}]$ is a $\mathcal{G}$-measurable integrable function for which
$$ \int_{E} X \, d\Bbb{P} = \int_{E} \Bbb{E}[X \mid \mathcal{G}] \, d\Bbb{P} \qquad \forall E \in \mathcal{G} $$
is true.

If $E$ is $\sigma(B)$-measurable, then for all $F \in \sigma(B)$ we have
\begin{align*}
\int_{F} \Bbb{E}[A\mathbf{1}_E \mid B] \, d\Bbb{P}
&= \int_{F} A\mathbf{1}_E \, d\Bbb{P} \qquad & \text{(by definition with $A\mathbf{1}_E$)}\\
&= \int_{F\cap E} A \, d\Bbb{P} \\
&= \int_{F\cap E} \Bbb{E}[A \mid B] \, d\Bbb{P} & \text{(by definition with $A$)} \\
&= \int_{F} \mathbf{1}_E \Bbb{E}[A \mid B] \, d\Bbb{P}
\end{align*}
and hence $\Bbb{P}$-a.s. $\Bbb{E}[A\mathbf{1}_E \mid B] = \mathbf{1}_E \Bbb{E}[A \mid B]$ holds.
Now you may invoke the standard mechanism - the monotone class theorem - to check that the same is true for all $\sigma(B)$-measurable r.v.s $X$ for which $AX \in L^1(\Bbb{P})$.
Alternatively, approximate $B$ by a sequence of simple functions and use the observation above directly together with an appropriate convergence theorem.
(Either cases, you may need to invoke conditional version of MCT or DCT.)
A: Definition: $$E[AB\mid B]$$ is any random variable $Z$ s.t.


*

*$Z$ is $B-$measurable

*$\forall B_1 \in \sigma(B)$
$$\int_{B_1} Z \, d\mathbb P = \int_{B_1} E[AB\mid B] \, d\mathbb P$$
or
$$E[Z1_{B_1}] = E[E[AB\mid B]1_{B_1}]$$

Now we must check if $BE[A\mid B]$ satisfies those.


*

*$Z = BE[A|B]$is $B-$measurable


because $B$ is $B$-measurable, $E[A\mid B]$ is $B-$ measurable and the product of $B-$ measurable functions is $B-$ measurable.


*$\forall B_1 \in \sigma(B)$


$$LHS = E[Z1_{B_1}] = E[BE[A\mid B]1_{B_1}] = E[BE[A1_{B_1}\mid B]] = E[E[BA1_{B_1}\mid B]] = E[BA1_{B_1}]$$
$$RHS = E[E[AB\mid B]1_{B_1}] = E[E[AB1_{B_1} \mid B]] = E[AB1_{B_1}]$$
QED

Observe that we used properties of $E[A\mid B]$:


*

*$E[A\mid B]$ is $B$-measurable

*$$E[A\mid B]1_{B_1} = E[A1_{B_1}\mid B]$$
A: To avoid questions of existence of these expectations, let's assume $A$ and $B$ are bounded a.s.
One definition of conditional expectation: 
$\mathbb E[X\mid B]$ is a measurable function of $B$ such that $\mathbb E[ g(B) \mathbb E[X\mid B]] = \mathbb E[g(B) X]$ for all bounded measurable $g$.
Any two of these are equal a.s. 
So let's see: $B \; \mathbb E[A\mid B]$ is a measurable function of $B$, and 
$$ \mathbb E[g(B) B \; \mathbb E[A\mid B]] = \mathbb E[g(B) B A] $$ 
